A proposal will make years that end in double zeroes a leap year only if the year leaves a remainder of 200 or 600 when divided by 900. Under this proposal, how many leap years will there be that end in double zeroes between 1996 and 4096?
Solution: We start with 1800, a multiple of 900, and add 200 to get 2000. So 2000 has a remainder of 200 when divided by 900. The next year with a remainder of 200 when divided by 900 is $2000+900=2900$. The year after that is $2900+900=3800$. Adding another 900 would result in a year greater than 4096.
Now we add 600 to 1800 and get 2400, which has a remainder of 600 when divided by 900. The next year with a remainder of 600 is $2400+900=3300$. Adding another 900 would result in a year greater than 4096.
So the years with remainders of 200 or 600 are 2000, 2900, 3800, 2400, and 3300. All of them end in double zeroes, so they are all leap years. We have a total of $\boxed{5}$ leap years.

OR

We can create inequalities. A leap year is equal to either $900a+200$ or $900b+600$, where $a$ and $b$ are positive integers. We solve for how many possible values of $a$ and $b$ we have. $$1996<900a+200<4096\qquad\Rightarrow 1796<900a<3896$$ So the value of $a$ can be 2, 3, or 4, giving us 3 different leap years. $$1996<900a+600<4096\qquad\Rightarrow 1396<900b<3496$$ So the value of $b$ can be 2 or 3, giving us 2 different leap years. In total we have $\boxed{5}$ leap years.


OR

We will end up with leap years when we add 200 or 600 to a multiple of 900. With 1800, we can add 200 or 600 to get two leap years. With 2700, we can add 200 or 600 to get two leap years. With 3600, we only get one leap year since $3600+600=4200$ is after 4096. We get a total of $\boxed{5}$ leap years.